$X$, $Y$ homeomorphic compact Hausdorff spaces then $C(X)$ and $C(Y)$ are $\ast$-isomorphic unital $C^\ast$-algebras Theorem: Suppose $\mathscr{A}$ is a singly generated, commutative, unital $C^\ast$-algebra with $\mathscr{A}=C^\ast(A)$ for some $A$ which is necessarily normal. There is a unique $\ast$-isomorphism of $\mathscr{A}$ onto $C(\sigma(A))$ mapping $A$ to the identity function on $\sigma(A)$.

Claim: If $X$ and $Y$ are homeomorphic compact Hausdorff spaces, then $C(X)$ and $C(Y)$ are $\ast$-isomorphic unital $C^\ast$-algebras.

How can this happen? Is it related to https://math.stackexchange.com/a/268025/593877?
Thanks in advance.
 A: We know that $C(X)$, $C(Y)$ are unital $C^*$-algebras, which doesn't have anything to do with $X$ and $Y$ being homeomorphic. So I won't elaborate on that part.
Following the proof in your link: Suppose that $h:Y\to X$ is a homeomorphism. Define $H:C(X)\to C(Y)$ by $H(f)=f\circ h$. This is a continuous, scalar valued function on $Y$. It is continuous because it is a composition of continuous functions.
Fix $y\in Y$, $f,g\in C(X)$, and scalars $a,b$. Then $$H(af+bg)(y)=(af+bg)(y)=af(y)+bg(y)=aHf(y)+bHg(y),$$ so $H$ is linear.
We also have that for any $f,g\in C(X)$, $$[H(f)H(g)](y)=f(h(y))g(h(y))=(fg)(h(y))=H(fg)(y),$$ so $H(fg)=H(f)H(g)$.
Moreover, $$H(\overline{f})(y)=\overline{f}(h(y))=\overline{f(h(y))}=\overline{Hf(y)},$$ so $H(\overline{f})=\overline{Hf}$. This shows that $H$ is a $*$-homomorphism.
We note that $H$ is an isomorphism because its inverse $H^{-1}:C(Y)\to C(X)$ is given by $H^{-1}g=g\circ h^{-1}$. To see that this is the inverse of $H$, we note that $HH^{-1}g=g\circ h^{-1}\circ h=g$ and $H^{-1}Hf=f\circ h\circ h^{-1}=f$.
This is also an isometry. Because $h$ is a surjection, $f$ and $f\circ h$ have the same range.
